Problem 1:
Answer1

The next BEST move for A would be D. If A is maximizing player then the next level would be minimum and then the next would be maximum and so on.
MAX:
MAX(E)=Minmax(L,M)=MINMAX(RESULT (7,8))=7
MAX(F)=Minmax(N,O)=MINMAX(RESULT (5,8))=8
MAX(G)=Minmax(P,Q)=MINMAX(RESULT (2,3))=3
MAX(H)=Minmax(R,S)=MINMAX(RESULT (0,-2))=0
MAX(I)=Minmax(T,U)=MINMAX(RESULT (6,2))=6 MAX(J)=Minmax(V,W)=MINMAX(RESULT (5,8))=8
MAX(K)=Minmax(X,Y)=MINMAX(RESULT (9,2))=9

MIN:
MIN(B)=MINMAX(E,F,G)=MINMAX(7,8,3)=3
MIN(C)=MINMAX(H,I)=MINMAX(0,6)=0
MIN(D)=MINMAX(J,K)=MINMAX(8,9)=8
MAX:
MAX(A)=Minmax(A,B,C)=MINMAX(RESULT (3,0,8))=8

Answer 2

Nodes that should not be examined while alpha beta pruning are:
Node O, node P, node Q, node T ,node U, node Y

Initially at
Node A: v=-∞ (since A is max) then it traverses to Node B where(α,β)=( -∞ , +∞ ).
Node B: v=+∞ (since B is min) then it traverses to Node E where(α,β)=( -∞ , +∞ ).
Node E: v=-∞ (since E is max) then it traverses to leaf Node L where(α,β)=( 7 , +∞ ) (since 7 is better than -∞) leaf Node M where(α,β)=( 7 , +∞ ) (since 7 is better than 6) so at Node E: v=7
In the already explored path at Node B: v=7 where(α,β)=( -∞ , +∞ )
Node F: v=+∞ then it traverses to
Leaf Node N where(α,β)=( 8 , +∞ ) (since 8 is better than 7)
Here the maximizer compares the value with other alpha values.
O,P, Q ,T ,U and Y nodes need not be expanded as the alpha-beta pruning algorithm is an alternative which explores a branch only if there is a possibility of producing superior alternatives.

Answer 3
MINIMAX procedure is a depth first search as one starts at the root (selecting some arbitrary node as the root) and explores as far as possible along each branch before backtracking.

Problem 2
Answer1
Variables
The dimension of the rectangular room are in terms of X and Ycoordinates.
So, the function would be f: {(X,Y)}
As for the rectangular tiles which have dimensions in the form of (x1, y1) and (x2, y2) coordinates, we can make use of the function f1:{( x1, y1):(x2, y2)
Domain (D)
The possible values of (x1,y1) & (x2,y2) which can be from {(0,0),(0,1),(1,0), ------, (X,Y)}
We make sure that X!=Y

Constraints:
C1:Size of room is given as X*Y.
C2: There should be no overlapping.
C3: No.of rectangle that can fit in the room is: f:{(X,Y)} / f1: {( x1, y1):( x2, y2)}

Answer 2
Variables:
Subject {Vs1,Vs2,…,Vsi}
Classroom available {VR1,VR2,…,VRj}
Time {T1,T2,……,Tk}

Domains:
DR = {rooms}
DTk = {time slots}
DC = {professors who teach subjects}

Constraints

C1: Let the number of Professor be A
C2: Number of available room be B
C3: Maximum no of class taught by professor in a day C
C4: No professor takes 2 classes at same time
C5: One hour class in morning
C6: No class in afternoon

Answer 3

A1 A2 A6 A3 A7 A5 A4
Suppose X1, X2, X3, X4 and X5 are the 5 available cities. The distance covered while travelling from one city to the other is as follows:

X1 – X2 = A1
X2 – X3 = A2
X3 – X4 = A3
X4 – X5 = A4
X5 – X1 = A5
X1 – X4 = A6
X3 – X5 = A7 all the cities must be visited at least once at least and the cost of travelling should be minimum the possible solution

X1 –> X2 ≠ X2 –> X1 This is because the travelling from X1 – X2 ≠ X2 – X1, as we need to travel a city only once X2 -> X3 ≠ X3 -> X2
Similarly x2->x3 ≠ x3->x2,as the city has to be travelled only once. X3 –> X4 ≠ X4 –> X3 , X4 –> X5 ≠ X5 –> X4, X5 –> X1 ≠ X1 –> X5

Problem 3

+
S

M E

O N

R D

E
S 9
E 5
N
6
D 7
M 1
O 0
R 8
Y 2
CASE: When M=1
M O N E Y

=>When M=1, S =9( since there S+1=1O) S has to be a digit which when added with M gives a result more than 10. The maximum digit we can allot S is 9. Therefore, S=9.
=> By mix and match we try all the possibilities for E. Let E=5 and given that M=0, (we know that E!=N), therefore there has to be 1 as carry over. This gives us N=6.
=>Now we have 6+R=5 which is greater than 10. R has to be some digit which when added gives 15.So R cannot be =9.Therefore there is 1 as carry over .so R=8.
=> Now we have D+5=Y, we are left out with didgits 2,7,3,4. D =7 as it has to occupie the maximum digit which gives a carry over to the next one. Therefore automatically Y=2.
Problem 4:
Answer1:
I am standing at Green House and all houses are in circle. Sequence from Green to left is Ivory, Blue, Red, Yellow, Green, the solution is :
ZEBRA stays in YELLOW House and in IVORY House they drink water
COLOUR IVORY RED YELLOW BLUE GREEN
NATIONALITY NORWEGIAN ENGLISH UKRANIAN SPANIARD JAPANESE
CANDY SMARTIES SNICKERS KITKAT HERSHEYS MILKIWAYS
DRINK WATER ORANGE TEA MILK COFFEE
PET SNAIL FOX ZEBRA DOG HORSE

Answer2:
I am standing at Green House the solution is: ZEBRA stays in GREEN House and in IVORY House they drink water. COLOUR YELLOW BLUE IVORY GREEN RED
NATIONALITY UKRANIAN SPANIARD NORWEGIAN JAPANESE ENGLISH
CANDY KITKAT HERSHEYS SMARTIES MILKIWAYS SNICKERS
DRINK TEA MILK WATER COFFEE ORANGE
PET FOX DOG SNAIL ZEBRA HORSE